## Orientation of Hybrid Orbitals

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

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### Orientation of Hybrid Orbitals

In the first example of today's lecture regarding the hybridization of ammonia, Dr. Lavelle says that we need to have sp3 hybridization because a bonding model with just the normal s and p orbitals would mean that the three unpaired electrons in the p orbital would have 90 degree bond angles, which is not correct. Is the reason they would all have 90 degree bond angles because of the orientation of the p orbitals? Does that mean that hybridization alters the orientation of the orbitals? I am just trying to understand what hybridization means in a physical sense.

Zaid Bustami 1B
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### Re: Orientation of Hybrid Orbitals

Hey Stuti!
Yea you're absolutely right. p orbitals in an atom are oriented 90 degrees from one another, so in molecules like ammonia or methane the orbitals have to hybridize in order for the 109.5 degree bond angles to be possible. I have a picture here that visualizes what happens when carbon orbitals hybridize in methane:

If you think about it, the hybridization of orbitals is what makes the tetrahedral arrangement of electron density in ammonia and methane possible. Prior to hybridization, the electron density can only be arranged around the atom as dictated by the p and s orbitals, but once these orbitals hybridize you're left with four identical hybrid orbitals that will want to move as far away apart as possible, lending to that tetrahedral shape. So in a spatial sense, hybridization allows for the creation of more "flexible" regions of electron density that behave as you would expect using the VSEPR model. This is why Prof. Lavelle keeps telling us to base hybridization on the VSEPR shape, since the hybridization is what enables the VSEPR shape to be possible.